Question

(x+1)(x+2)(x+3)(x+4)=120

Answer 1

(x+1)(x+2)(x+3)(x+4)=120

=>(x+1)(x+4)(x+2)(x+3)=120

=>{x²+5x+4}{x²+5x+6}=120

Let x²+5x+4=A

=>A(A+2)=120

=>A²+2A=120

=>A²+2A-120=0

=>A²+12A-10A-120=0

=>A(A+12)-10(A+12)=0

=>(A-10)(A+12)=0

A=12,-10

WHEN A=12

x²+5x+4=12

=>x²+5x-8=0

x=-5±√(5)²-4(1)-8)/2×1

x=-5±√1/2

When a=-10

x²+5x+4=-10

=>x²+5x+14=0

x=-5±√(5)²-4(1)(14)/2

x=-5±√-31/2

Answer 2

120=(x+1)(x+2)(x+3)(x+4)=((x+1)(x+4))((x+2)(x+3))120=(x+1)(x+2)(x+3)(x+4)=((x+1)(x+4))((x+2)(x+3))

=((x2+5x+5)−1)((x2+5x+5)+1)…(1)=((x2+5x+5)−1)((x2+5x+5)+1)…(1)

=(x2+5x+5)2−1=(x2+5x+5)2−1,

we get x2+5x+5=±11x2+5x+5=±11.

Thus x2+5x−6=0x2+5x−6=0 or x2+5x+16=0x2+5x+16=0. The first of these give x=1x=1 or x=−6x=−6. The second of these give x=12(−5±−39−−−−√)x=12(−5±−39).

It is easy to verify that x=1x=1 and x=−6x=−6 are both solutions; in fact, 120=2⋅3⋅4⋅5=(−5)⋅(−4)⋅(−3)⋅(−2)120=2⋅3⋅4⋅5=(−5)⋅(−4)⋅(−3)⋅(−2).

That the second pair of complex conjugates is also a solution can be best verified by replacing x2+5xx2+5x by −16−16 in eqn. (1)(1), say.

There are two real solutions, x=1x=1 and x=−6x=−6, and two non-real solutions x=12(−5±−39−−−−√)

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